Transcript Odds - PBworks

MDM4UI
Unit 3 - Lesson 4
Probabilities Using
Counting Techniques
This section examines methods for determining the
theoretical probabilities of successive or multiple events.
P(A) = n(A)
n(S)
EXAMPLE 1:
Two brothers enter a race with five friends. The racers draw lots to
determine their starting positions. What is the probability that the
older brother will start in lane 1 with his brother beside him in lane 2?
n( S ) 
7
n( A) 
P2
n( A)
P( A) 
n( S )
1

42
 2.3%
1
EXAMPLE 2:
A focus group of three members is to be randomly
selected from a medical team consisting of five
doctors and seven technicians.
(a)
What is the probability that the focus group
will be comprised of doctors only?
(b)
What is the probability that the focus group
will not be comprised of doctors only?
SOLUTION
(a)
What is the probability that the focus group
will be comprised of doctors only?
n( A) 
5
C3  10
n(S )  12 C3  220
 the probability of selecting a focus group of only doctors is
n( A)
n( S )
10

220
1

22
P( A) 
 0.045  4.5%
(b)
What is the probability that the focus group
will not be comprised of doctors only?
 this question is asking for the probability of the
complement of A which we denote P(A').
P( A ')  1  P( A ')
1
 1
22
21

22
 0.955
 95.5%
The Birthday Test
How many students in the lab today have the same birthday?
EXAMPLE 3:
What is the probability that two or more students
out of a class of 24 will have the same birthday?
Assume that no students were born on February
29.
SOLUTION
The simplest method is to find the probability of the
complementary event that NO TWO people have the
same birthday.
P(two students have different birthdays) 
364
365
P(three students have different birthdays)  364  363
365 365
If we continue this process, we get ...
P(24 students have different birthdays) 
364 363 362



365 365 365

342

365
P( A ') 
0.462
 P( A)  1  P( A ')  1  0.462  0.538
Homework
for this lesson
p. 324 # 5, 8, 9, 11, 12.